The Square Root of n!
Date: 10/14/98 at 08:14:43 From: preston Johnson Subject: Algebra, square roots, and factorials For what natural numbers n is the square root of n! an integer? I tried putting different numbers in for n. If I did the problem at all right, 1, 23, and 24 all work. I do not know how to go about this problem or how to set up an equation. If you could help, that would be great.
Date: 10/14/98 at 09:30:41 From: Doctor Mitteldorf Subject: Re: Algebra, square roots, and factorials Dear Preston, Among factorials, only 0! and 1! are perfect squares. Your result for 23 and 24 is probably due to the round-off error in your calculator or computer. To prove that a factorial bigger than 1 can't be a perfect square, first think about breaking down the factorial into prime factors. For example, 4! = 24, and 24 = 2*2*2*3. In order for any number to be a perfect square, it must contain an even number of each prime factor. Now think about the largest prime less than or equal to n. Call that number p. That number appears only once in the list of factors 1*2*3*4*5*...*p*...*n, so it can't appear an odd number of times unless 2*p is also less than n. Now there's a semi-famous theorem called Bertrand's Postulate, which says that there is always a prime number between n and 2n. This is just what you need to show that 2p isn't less than n - because if it were, there would have to be a larger prime than p that is also less than n. The proof of Bertrand's postulate isn't easy, but you can read it at: http://mathforum.org/dr.math/problems/bolton.8.1.96.html You may also be interested in programming in a language that allows you to work with numbers that have more than 20 or so digits, which most calculators and computer programing languages can't do. Look at the program Mathematica - it contains wonders! - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
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